Mathematics - 2017-04-17 (page 2 of 6)

Balarka Sen

03:00

sort of

Daminark

Also it depends a lot on the professor

Adeek

I am trying to understand the a step in the following proof I just want to make sure my reasoning is correct

Daminark

Like, at least in my school, the curriculum is not set in stone at all

Adeek

@BalarkaSen share.pho.to/AflDh Consider the following

Each element of G acts as a deck transformation and the covering space is normal since g_2g_1^{-1} takes g_1(U) to g_2(U)

so what I am trying to understand is normality issue so we have that by Y/G what we mean is actually (Y/G,G)

so now if we take the fibre over G is of the form of g_i(U)

so is why fibers are of the form g_i(U) ?

Is that correct @BalarkaSen ?

Daminark

Like there are vague standards, analysis is supposed to cover the basics up to stuff like metric spaces and uniform convergence, and then linear algebra, multivariable calculus, Lebesgue theory, and functional analysis. Beyond that you have leeway

One professor used to make it heavy on p-adics and Fourier analysis on groups

Excalibur42

03:04

fourier analysis on groups what?

Daminark

With the professors now it does stuff like differential forms, some ODE, and Marianna likes to do a bit of basic geometric measure theory and topological games

Balarka Sen

@Adeek Actually I'm a little busy right now. maybe someone else can help

Adeek

alright

I will ask on main I guess

Daminark

So you can give a topology to a group which "plays nicely" with its operation

Adeek

yeah

@Daminark many covering spaces arise this way

Daminark

03:07

Meaning that the operation is a continuous function on $G^2$, and the map from an element to its inverse is also continuous on $G$. Then you can define something called the Haar measure which is compatible with this, and talk about $L^2(G)$ and do Fourier analysis. Turns out, when you have locally compact abelian groups, there's a nice theory to handle this, I think called Pontryagin duality or something

(I know little of this, only vaguely)

Oh that's neat @Adeek!

What... are covering spaces?

Adeek

@Daminark you can think of them as subgroups of $\pi_1(X,x_0)$

But there is a geometric description

Excalibur42

That sounds pretty cool actually

Daminark

First homotopy group of $X$ centered at a base $x_0$?

Adeek

yeah

Daminark

(Meaning fundamental group lol)

(I'm slow)

Victor

03:09

Anybody can solve this challenging junior high math problem with other method ?
https://math.stackexchange.com/questions/2237789/what-is-an-alternative-method-either-advance-intermediate-or-beginning-for-solv

thanks in advance

Adeek

So geometrically they are locally copies of the original space

in some strong sense

Daminark

Hmm, that's interesting

Adeek

There is a correspondence between subgroups of the fundmental group and a particular unique covering space.

Daminark

Nice

Adeek

yeah its elegant

Daminark

03:21

Kinda funny how aside from the normal stuff like fields and vector spaces that people do in analysis and prior, the main groups I've dealt with are finite groups

Victor

@Daminark look like you are interested in abstract algebra, is there a way to solve my problem with an abstract algebra alternative way?

Daminark

I can try for sure!

But note that I never really took the class, I just kinda did a bit of group theory for a Sylow paper and blasted through a book in order to (not really but almost) get some background for a class on algorithms in finite groups

So I'm kinda inexperienced. Still, it should be fun!

Excalibur42

@Victor looks like Cardan's formula for the cubic at a glance

Daminark

Yeah there's no clean way of doing it, it seems

Like you can't just say "By the Sylow theorems we get _____"

@DHMO Sets are not doors

DHMO

@Daminark ?

Daminark

03:30

Your comment on the $E^1$ post

DHMO

I know what I commented

what do you want to say?

Daminark

I just found that amusing

DHMO

@Daminark do you know why I said that?

Daminark

I assumed it was a pun on sets being open or closed?

DHMO

More than that

I don't just comment randomly

Excalibur42

03:33

Whoa cauchy sequences of equivalence classes of cauchy sequences is doing my head in

DHMO

@Excalibur42 what the hell?

Daminark

What is it then?

Excalibur42

Closure of a metric space by defining an equivalence relation between Cauchy sequences

Daminark

@Excalibur That sounds like saying completing a metric space?

Yeah

Excalibur42

yep

DHMO

03:35

@Daminark because (a) is asking whether a set is open

and OP said that the set is closed

Daminark

Oh, so you were going for the whole, closed and open not being mutually exclusive

DHMO

ya

Excalibur42

Yeah theres this step: Let $(\overline{x}_n)$ be Cauchy, then given $\epsilon >0$ there exists $N_1$ such that $n,m \geq N_1$ implies $D(\overline{x_n},\overline{x_m})<\epsilon$, where $D(\overline{x},\overline{y})$ is defined by $\lim_{n \rightarrow \infty}d(x_n,y_n)$.

Daminark

Lol, some people might not pick up on this, I didn't realize you were addressing that mistake, I'm just like "Wait no it's totally open because intervals, also completely unrelated kek that's funny DHMO"

DHMO

@Daminark and I don't care about those people

Excalibur42

03:38

but I don't get why $\lim d(x_n,y_n)$ implies there is a $N_2$ such that $k \geq N_2$ implies $d(x_k^n,x_k^m)<\epsilon$

Daminark

I mean, for what it's worth, that group of people is likely to include the OP

Will Njundong

hey guys: imgur.com/a/HMh2Z

I dont know exactly what im being asked to do

DHMO

@Daminark then you can help me clarify

Excalibur42

way to many indices to even bother writing out

DHMO

(I don't quite care about the OP either)

@WillNjundong $\displaystyle \int_0^x 7 \ \mathrm dt = 7x$

@Daminark you didn't address (a)

Will Njundong

03:41

@DHMO i see thank you

Excalibur42

@Daminark imgur.com/pg3jrq2 send help

just that last sentence

Daminark

I might just write an answer outright in the case of (a), it seems like the answer given was just random

DHMO

@Daminark just add "just like what you did in (a)" to your comment

Daminark

@Excalibur42 Elements in the Cauchy sequence are equivalence classes of Cauchy sequences. So saying that $d(x_n,x_m)$ is small, you're saying (by how you defined the equivalence classes) that $d(x_n^k,x_m^k)$ is small for whatever representatives you chose

And did so @DHMO

DHMO

@Daminark thanks

Excalibur42

03:48

.......yep okay I think I got it

Daminark

No problem!

And awesome!

Excalibur42

Whoa why does showing that any $\overline{x} \in \overline{X}$ is the limit of a sequence in $\varphi(X)$, where $\varphi$ identifies elements of $X$ with constant sequences, give that $\varphi(X)$ is dense in $\overline{X}$?

Daminark

Another characterization of density is that $A$ is dense in $B$ if for any $b\in B$, there exists a sequence in $A$ converging to it

Because the limit of a sequence in $A$ is either a point in $A$ or a limit point of $A$

Excalibur42

ah...okay that seems actually trivial when you put it that way

Daminark

Yeah there are a lot of concepts which aren't intrinsically hard, but the presentation can make them sound like it

Hey @s.harp and @Tim

Tim The Enchanter

04:02

@Daminark Hey Dam.

Daminark

How's it going?

Tim The Enchanter

@Daminark Pretty good, I've read through about half of last nights messages before I got fed up. But I've got tea so I'm happy.

You?

Daminark

Finishing up the analysis pset (last problem is to show that $\mathbb{N}^{\mathbb{N}}$ is homeomorphic to $\mathbb{R}\setminus\mathbb{Q}$)

DHMO

@Daminark what the hell

how are they homeomorphic?

Daminark

It's a bit of a surprise, the vague idea is to sort of take a bunch of open intervals that exclude the integers, and then find subintervals of each and keep this construction going

DHMO

04:15

Wait, what is the topology on $\Bbb N^\Bbb N$?

Daminark

Discrete and then product

Which is equivalent to the cylinder topology on the countable game

Will Njundong

imgur.com/a/29Q60

can someone explain where they got $77/2$ here?

Daminark

Basically, a branch of the game is a point, and you define a base for the topology by way of looking at all the branches that are diverge at some spot

DHMO

@WillNjundong you need to add the contribution from the previous section

@Daminark I know nothing about game theory

@WillNjundong or you can just use the area of trapezium

Daminark

It's not game theory in the sense that you see in economics, or surreal numbers a la Conway (though those are cool nonetheless)

Will Njundong

04:21

@DHMO but then Im still confused as to how they got 77

Tim The Enchanter

Well damn. I just realised a homeomorphism and an homomorphism are not the same thing. I am in way over my head

Daminark

The idea is that you just model a game as just the set of decisions, so it works like a tree

DHMO

@WillNjundong 0.5[7][(x-2)+(x-9)] = 7x-77/2

Daminark

So as an example, a game where each player has 2 choices can be modeled by the binary tree

Which, under the game topology, is homeomorphic to the Cantor set

DHMO

god

Daminark

04:24

So let's say you generate the topology on some game, you can talk about Borel sets

Tim The Enchanter

Is $\mathbb{N}^{\mathbb{N}}$ countable? Seems like it should be.

DHMO

@TimTheEnchanter no it isn't

and it is a common misconception

Daminark

@Tim $2^{\mathbb{N}}$ is the set of sequences of 0 and 1, which isn't countable

But yeah, the cool thing is that given a game (2 player), if the set of branches that lead a player to victory is Borel, then someone has a winning strategy

Tim The Enchanter

Wait so binary trees are uncountable as well then?

DHMO

@TimTheEnchanter correct

Mike Miller

04:27

infinite binary trees, note

Daminark

We didn't actually go for the proof of that statement, it's kinda heavy (would've taken us quite a few lectures) and we were mostly concerned with using Banach-Mazur to prove things, such as that the typical continuous function is not monotone in any interval

Hey @Mike!

Mike Miller

hi

Daminark

How's it going?

Mike Miller

you're so constantly excited

i'm writing, so a solid feh

Tim The Enchanter

@MikeMiller Ah yes

Daminark

04:31

Haha, that's pretty true. I dunno, excitedly mutters something vaguely justifying this state

Hey @Balarka!

Balarka Sen

yo

Daminark

What's up?

Tim The Enchanter

Wait so if I construct a map from $P(\mathbb{N}) $ to the set of all sequences of 0 and 1, as $ \forall A \subset \mathbb N $, $\space A \mapsto { a_n}$ $\space, a_n = 1$, $n \in A ,\space a_n = 0$ , $n \notin A$, that would be a bijection?

Daminark

04:48

Yup

Tim The Enchanter

And if an operation on the first set was $A \cup B - A \cap B$ and the operation on the second was XOR'ing the sequences, that would be a homomorphism under these relations?

Or do the two sets mapped by the homomorphism need the same operation?

Alessandro Codenotti

05:04

Hi chat

DHMO

@AlessandroCodenotti buongiorno

Alessandro Codenotti

Thanks for the link @Balarka

I should really learn about homology

Martin Sleziak

@TimTheEnchanter That would be a group homomorphism. You can get even ring homomorphism, namely $(P(X), \triangle, \cap)$ is a ring homeomorphic to $(\mathbb Z_2)^X, \oplus, \odot)$. I guess there are a few posts on the main site about this.

user97303

hey

user97303

Suppose you have span{v} and span{w} over GF(q), where v and w are vectors

Martin Sleziak

05:14

I have mentioned this in my answer here. But I guess there are several other posts mentioning this correspondence.

user97303

is it still true that if $span{v} \cap \span{w}$ is nonempty then v is in span{w}?

Daminark

@DHMO Turns out, this space is a thing in descriptive set theory, called "The Baire Space"

DHMO

@Daminark what space

Daminark

$\mathbb{N}^{\mathbb{N}}$

DHMO

I see

user97303

05:16

this is a really basic question but I only ever learned about lin alg over the reals and complex

Martin Sleziak

@chell Well, you always have zero vector in the intersection. You probably meant "non-trivial" rather than "non-empty". If you meant this, it is true in any vector space over arbitrary field.

user97303

Yeah, that's what I meant, thank you

Balarka Sen

@AlessandroCodenotti I advocate learning homology.

Daminark

There is a theorem saying that this is the unique non-empty, Polish, 0-dimensional space such that any compact subsets have empty interior

Tim The Enchanter

@AlessandroCodenotti Hey Alessandro

Daminark

05:17

By Alexandrov and Urysohn

Tim The Enchanter

@MartinSleziak Ah thanks Martin

Daminark

There's another by Brouwer stating that the Cantor space is the unique perfect, non-empty, compact metrizable, zero dimensional space

(All this up to homeomorphism)

Martin Sleziak

@chell If you have $x\ne0$ which belongs to both spans, then $x=cv=dw$ for some scalars c, d. So you get $v=c^{-1}dw$ and $v\in\operatorname{span}(w)$.

To make things a bit more confusing, there is a difference between the Baire space and a Baire space.

Daminark

Yeah, we had talked about that as well

Martin Sleziak

However, the Baire space is a Baire space.

Daminark

05:21

Actually in class we made it even worse

Because a set can also have the property of Baire

user97303

It just seemed very unintuitive to me because I was trying out examples and it seems like this might not be true. Say over $GF(3)$, $2(1,2) = (2, 1)$ so you can get $(1,1)$, $(2,2)$, $(0,2)$, $(0,1)$ in the span

Daminark

I forget the exact condition, I believe it was equivalent to having an associated winning strategy in Banach-Mazur

Something about its symmetric difference with some open set being first category or whatever

This fact didn't come up too much though

Martin Sleziak

@chell If you have a field with three elements, then span of single vectors consists of 3 vectors. For example $(1,1)$ is not in the span of $(2,1)$ if that's what you meant there.

Since you have some questions related to linear algebra, I should mention that there also is Linear & Abstract algebra. But I have to admit that it is not very active.

user97303

Oh nevermind, I thought $(1, 1)$ is the difference between $(1, 2)$ and $(2, 1)$. Wow I'm a derp.

user97303

calculations are hard :(

Koolman

07:01

0

Q: How many Digits in $3^{2020}$

I know n+1 = number of digits in $3^{2020}$ but how to find number of digits . How can we do it without using a calculator .

Anonymous

07:21

@Koolman You need to remember the value of log(3). That's it.

Anonymous

I don't think there is any simpler method than that.

Anonymous

Or there could be a hack using binomial theorem

Anonymous

$(9)^{1010}=(1-10)^{1010}$

Studentmath

So, I've got a question

Balarka Sen

You could prob also use the Taylor series. $\log(1-x) = -\sum x^k/k$; plug in $x= 2/3$. That converges pretty fast.

(don't use $\log(1+x)$!!! that's alternating and would take ages to converge)

Studentmath

07:27

Don't think he can use logs or taylor

(Because of the limitations on the allowed methods)

Balarka Sen

Ah, no calculators. Oh well!

Studentmath

Yeah, I never liked these kind of questions

Daminark

Rip

Balarka Sen

Well if it's multiple choice then the $9^{1010}$ tricks like that would tell you something.

Daminark

Plot twist: The choices are all 1 away from each other

Studentmath

07:30

Haha

Balarka Sen

well, you're probably screwed then

Daminark

Rehi @Mike!

But yeah, they prob had far away stuff, like doing this without a calculator is going to be one of the vague estimate type problems

You know, how $\pi^2 = e^2 = 3^2 = 10$, that kind of thing

Koolman

@blue i don't think this hack will work here

@BalarkaSen you know jee does not allow calculators

Balarka Sen

I don't.

Anonymous

Hush. Don't mention that word here, koolman =P

Koolman

07:34

fine

Anonymous

Soon 5 people will pounce on you asking for your phone number

Anonymous

lol

Koolman

why

for whatsapp group ??

Balarka Sen

we shall meet again in India / as though we had buried education there / and we shall pronounce for the first time / the blessed word with no meaning ("JEE")

Studentmath

@Daminark Haha

Anonymous

07:35

@Koolman Well, I am thinking. If I get it I'll let you know :-)

SBM

Yes, JEE everywhere is like, really not that good of an idea

Secret

07:54

Old paper: Fractional cartesian product

Astyx

08:07

Hi chat

DHMO

@Astyx salut

SBM

Hello @Astyx

Secret

08:25

in The h Bar, 2 mins ago, by Slereah

A pretty picture said to be something from algebraic topology that I have no idea what it means

DHMO

@Secret each arrow is a group operation on the right

each point is a result

you start at the identity $e$

$a$ is of order $2$ and $b$ is of order $3$

as shown from the beautiful cycles

Secret

Ah I see

DHMO

it's literally infinite if you don't restrict it much

you can always go $ababababab...$

:36755691 I don't think $a$ and $b$ commute

Secret

ok

DHMO

@Secret do you think it can represent any group familiar to us?

Secret

08:30

@BAYMAX Seeing things in your mind's eye. If your brain don't allow you to see it, you will not see it. Thus one need to convince one's brain to let you perceive it (suddenly, all that esoteric advice make sense to me, lol)

@DHMO Well, I guess it will ran into problem for something like the tarski monster group, since there are too many prime elements

DHMO

08:48

@Secret what is a prime element?

Oh, every element is a prime

Secret

Sorry, I mean, an element that is of prime or prime powered order

DHMO

For example?

Secret

Tarski monster group

there's also another group in mind that is also quite large, but I need to recall its name, give me a sec

DHMO

Can you construct a subgroup of the group above?

@Secret I'm referring to this group

where $a^2=b^3=e$

Secret

Prüfer group

In mathematics, specifically in group theory, the Prüfer p-group or the p-quasicyclic group or p∞-group, Z(p∞), for a prime number p is the unique p-group in which every element has p different p-th roots. The Prüfer p-groups are countable abelian groups which are important in the classification of infinite abelian groups: they (along with the group of rational numbers) form the smallest building blocks of all divisible groups. The groups are named after Heinz Prüfer, a German mathematician of the early 20th century. == Constructions of Z(p∞) == The Prüfer p-group may be identified with t...

give me a sec when I construct your example...

Let's see...

Imposing additionally this relation $ab=ba=e=ab^2=a^2b$ should give us a finite subgroup

DHMO

08:57

$=$?

I doubt it is a subgroup. $(ab)(ba) = abba$ which is not in the set

Or did you mean to equate those five elements?

I thought equating elements is not allowed

Secret

O wait, sorry, you mean to find a subgroup, not to make one using the given group and change its structure

DHMO

yes, that is what I mean

I suspect that the only subgroup is $\langle a \rangle$ and $\langle b \rangle$

You know, any other subgroup generated from other elements go to infinity, intuitively

Secret

It's an infinite group, thus we don't have the nice decomposition theorem to say it is a direct product of finite p-groups. But yes, it definitely have the cyclic groups $\langle a\rangle\cong \Bbb{Z}_2$ and $\langle b\rangle\cong \Bbb{Z}_3$ as subgroups. Whether there are others I need to check...

Will Njundong

09:18

hi, is anyone on?

DHMO

@WillNjundong just ask

Mees de Vries

Good morning math chat. Would someone like to play a (mathematical) game with me?

DHMO

@MeesdeVries depends on what game it is

Mees de Vries

We start with an empty 3 x 3 matrix. We take turns adding one (let's say real) entry to the matrix. Player 1 wins if the matrix is invertible. Player 2 wins if it is not.

DHMO

what the hell

Will Njundong

09:20

find $G'(x)$ for $$G(x) = \int^x_3 5t dt$$

DHMO

this is tic-tac-toe kind of lol

Mees de Vries

I suppose!

DHMO

@WillNjundong use the fundamental theorem of calculus

Mees de Vries

Background: the game is trivial if the matrix has even dimensions; then player 2 can just copy any move player 1 makes in an even row, in the odd row above; and any move in an odd row, in the even row below.

DHMO

So we start with $\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$

Mees de Vries

09:21

It is also trivial for a 1 x 1 matrix (exercise for the reader).

Making 3 x 3 the first non-trivial case.

Yeah, I think *s might be better than 0s, though, because 0 is a valid move.

DHMO

right? @MeesdeVries

Secret

Round 1:
\begin{bmatrix}*&*&*\\*&*&*\\*&*&*\end{bmatrix}

Mees de Vries

Do you have a preference for turn order?

DHMO

not really

Mees de Vries

Alright. I might as well go, then:
$$
\begin{bmatrix}1&*&*\\*&*&*\\*&*&*\end{bmatrix}
$$

DHMO

09:24

heh

$$
\begin{bmatrix}1&-1&*\\*&*&*\\*&*&*\end{bmatrix}
$$

RIP LaTeX

Mees de Vries

$$\begin{bmatrix}1&-1&8\\ *&*&*\\ *&*&*\end{bmatrix}$$

Will Njundong

@DHMO Just looked at some solved examples and understood this theorem. that was easy XD . I dont go to class, so I had no idea what you were talking about lol.

DHMO

@WillNjundong google is your friend

Mees de Vries

I think I win once you play in the same row or column as me. Then I can complete that row or column, and since you have to start in the remaining $2 \times 3$ block, I win.

DHMO

well if I don't play in the same row, then you win immediately

Mees de Vries

09:28

How's that?

DHMO

how is that not

Mees de Vries

Uh, I don't know, that's why I'm asking :P

DHMO

because there are 6 diagonals

$$
\begin{bmatrix}1&-1&8\\ *&0&*\\ *&*&*\end{bmatrix}
$$

Secret

NB: I know nothing about the tools of game theory

DHMO

@Secret how is your search?

Vrouvrou

09:29

HELLO,is the metric space $(R,d)$ where $d(x,y)=|\exp(x)-\exp(y)|$ is complete ?

Mees de Vries

\begin{bmatrix}1&-1&8\\ *&0&*\\ *&5&*\end{bmatrix}

Six diagonals...?

DHMO

yes

because they can wrap

Secret

@DHMO Nah, nothing interesting I can find so far except any subgroups other than the cyclic ones, if it exists, must be infinite

Mees de Vries

OK, sure. But what does that have to do with playing in a different row?

DHMO

well if you play in a different row, then the two elements would be one of the six diagonals

wait

if i make it invertible now, do i win?

Mees de Vries

09:31

No, I'm trying to make it invertible.

DHMO

so we have to wait until the asterisks are filled?

wait what?

oh, I didn't read the rules carefully enough

Mees de Vries

Player 1 tries to make it invertible, player 2 tries to make it singular

DHMO

I thought we were trying to make it invertible lol

never mind

Mees de Vries

Haha

DHMO

i think you already won

due to my misunderstanding of the rules

$$
\begin{bmatrix}1&-1&8\\ *&0&*\\ -5&5&*\end{bmatrix}
$$

Mees de Vries

09:34

Do you want a rematch? :P

Secret

4

Q: Number of subgroups of an infinite group

Is there an infinite group with only a finite number of subgroups?

abstract-algebra group-theory

DHMO

@MeesdeVries sure

Secret

Yeah, in that case, the remaining ones will be the unions of the cyclic groups

Mees de Vries

(I would put a 2 in the left asterisk now, and then no matter what you put on either of the other asterisks I just make the next number something inconceivably large.)

Martin Sleziak

@N3buchadnezzar When dealing with duplicates, you might also use the dedicated chat room. And if everything else fails, there is also this thread on meta.

Vrouvrou

09:35

can someone help me please

Mees de Vries

Oh, hmm. How's this for a winning strategy: every number player 1 plays is much, much, much larger than all preceding numbers?

DHMO

@MeesdeVries just try

I think I can defeat you still

Mees de Vries

Alright. I'll repeat my first move from before.

DHMO

I mean

continue the game

Mees de Vries

Oh, huh. OK.

DHMO

09:36

so it's your turn

Mees de Vries

\begin{bmatrix}1&-1&8\\ 2&0&*\\ -5&5&*\end{bmatrix}

DHMO

and i win

$$
\begin{bmatrix}1&-1&8\\ 2&0&*\\ -5&5&-40\end{bmatrix}
$$

Because row 1 and row 3 are linearly dependent

Mees de Vries

\begin{bmatrix}1&-1&8\\ 2&0&10^{20}\\ -5&5&-40\end{bmatrix}?

DHMO

the determinant is still 0

BAYMAX

0

Q: Mapping of a horizontal line $y = c$ by $w = \frac{1}{z}$ onto which region?

I was trying to solve this problem , so $y = c$,suppose for ease let us take $c > 0$. so $w = \frac{1}{z} = \frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} = \frac{x - ic}{x^2 + c^2} = \frac{x}{x^2 + c^2} - i( \frac{c}{x^2 + c^2})$. Now as $w = u + iv$,comparing we get $u = \frac{x}{x^2 + c^2}$ a...

complex-analysis

DHMO

09:40

@MeesdeVries are you convinced?

Mees de Vries

Yeah. Huh. How about that.

I don't actually see the dependence moving before my eyes though.

DHMO

@MeesdeVries do you want a rematch?

Mees de Vries

Haha, sure.

Do you want to start now?

DHMO

doesn't matter

Mees de Vries

Because? You'll win either way? :p

DHMO

09:42

sure :p

so who starts?

Mees de Vries

You go.

DHMO

so I make it invertible now?

$$
\begin{bmatrix}1&*&*\\ *&*&*\\ *&*&*\end{bmatrix}
$$

Mees de Vries

Yeah.

\begin{bmatrix}1&-1&*\\ *&*&*\\ *&*&*\end{bmatrix}

DHMO

hey it's the same move lol

Mees de Vries

Yeah, I'm learning.

DHMO

09:45

$$
\begin{bmatrix}1&-1&*\\ *&*&2\\ *&*&*\end{bmatrix}
$$

Mees de Vries

Hmm.

\begin{bmatrix}1&-1&0\\ *&*&2\\ *&*&*\end{bmatrix}

DHMO

$$
\begin{bmatrix}1&-1&0\\ *&*&2\\ *&0&*\end{bmatrix}
$$

$0$ is not necessarily bad, I believe

Mees de Vries

I think you've won this one...? Then again, I thought I won the last one even when the matrix was completely filled.

DHMO

I am not seeing how I have won lol

@MeesdeVries are you here?

Mees de Vries

Yeah, I'm trying to think about whether I've already lost, seeing if I can derive a good next move from failing to show that. :-)

DHMO

09:54

sometimes one needs to take a leap of faith

@MeesdeVries it's been 8 minutes

Mees de Vries

Fine, fine.

DHMO

I apologize.

Mees de Vries

\begin{bmatrix}1&-1&0\\ 0&*&2\\ *&0&*\end{bmatrix}

Yeah, you're right. I'm more frustrated with myself than with the game. :-)

DHMO

lol

I see what you're trying to do

$$
\begin{bmatrix}1&-1&0\\ 0&*&2\\ 3&0&*\end{bmatrix}
$$

Mees de Vries

I genuinely don't really know. Like. This is an easy win if you put a 0 in any of the remaining spots, but I don't think you will.

DHMO

09:58

definitely linearly independent.

I think @TedShifrin or @BalarkaSen or @Semiclassical might be interested in this game... but we would have to wait for their active hours.

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$Mathematics$ Mathematics